MHB What is the probability of selecting two males given that both are grey?

[navi]

So I am having more problems with my homework...

For this problem, assume there are 6 grey females, 2 grey males, 6 white females, and 2 white males. Two mice are randomly selected. What is the probability of selecting two males given that both are grey?

So for selecting two males, I was thinking of 4/12, since there are 4 males in total out of 12. Then for the grey I was thinking of 8/12, since there are 8 grey mice. However, that did not work once I plugged it into the formula...

I thought of using combinations, but I am not sure how that would work and how I should write that. I was thinking of C(4,2)/C(12/2) for the intersection of grey mice and male, and for the "given that both are grey" C(8,2)/C(12,2)...?

Tbh, I am not sure of what I am doing...

 

[Ackbach]

So, conditional probability has the effect of "narrowing your world", so to speak. The two mice are both grey, so you can completely ignore the whites.

You pick the first grey mouse without replacement (I think that's understood, though it's good to be clear, because it does change probabilities). What's the probability that it's male? Now pick the second grey mouse, assuming that the first one you picked was male. What's the probability that this second one is male?

 

[HOI]

Given that both mice are gray then we can ignore the other mice. There are 6 gray females and 2 gray males for a total of 8 gray mice. The probability the first mouse drawn is male is 2/8= 1/4. There are then 6 gray females and 1 gray male. The probability the second mouse drawn is male is 1/7. Given that the two mice drawn are gray the probability they are male is (1/4)(1/7)= 1/28.